Question

sin7x - 2cos2x = \(\sqrt{3}\left(sin2x-cos7x\right)\)

Answer 1

Sửa đề: \(sin7x-cos2x=\sqrt{3}\left(sin2x-cos7x\right)\)

\(\Leftrightarrow sin7x\cdot\dfrac{1}{2}-\dfrac{1}{2}\cdot cos2x=\dfrac{\sqrt{3}}{2}\cdot sin2x-\dfrac{\sqrt{3}}{2}\cdot cos7x\)

\(\Leftrightarrow sin\left(7x+\dfrac{pi}{3}\right)=sin\left(2x+\dfrac{pi}{6}\right)\)

=>7x+pi/3=2x+pi/6+k2pi hoặc 7x+pi/3=-2x+5/6pi+k2pi

=>5x=-pi/6+k2pi hoặc 9x=1/2pi+k2pi

=>x=-pi/30+k2pi/5 hoặc x=1/18pi+k2pi/9